Some combinatorial properties of Fano geometry I am starting to learn about finite geometries and in particular am trying to understand some basic (mainly combinatorial) features of the seven point geometry.
(I have not yet reached the context of projective geometry, so my questions are truly basic, and I'd appreciate answers that use only elementary methods.)

Let's say that a symmetry is a permutation of $\{1, \ldots, 7\}$ that preserves lines. I am not sure how to prove whether or not the following properties hold:
(i) Given any two different lines (e.g., $\{1, 3, 2\}$ and $\{3,6,5\}$), there is a symmetry that takes one to the other.
(ii) Given any two distinct points, there is a symmetry taking one to the other.
Moreover, although I am able to see why there are 168 total symmetries, I can't see why there are 24 symmetries fixing a point and 24 fixing a line (as stated on WIkipedia).  I also can't figure out at all how many would fix two lines.
I'd really appreciate if someone could help me answer these questions. My intuition with finite geometry is pretty weak right now.
 A: To do calculations with the Fano plane it is convenient to work with a more natural model provided by homogeneous coordinates in $F^3$ where $F$ is the field with $2$ elements.  Here a point can be either thought of as a line through the origin, or the equivalence class of points in $F^3\setminus\{0\}$ under the relation of being proportional by a nonzero scalar in $F$. Then numbers of symmetries, etc. can be calculated using subgroups of the linear group over the field $F$.
A: For question (i) it suffices to show that for any line, there is a symmetry taking the line $\{1,2,3\}$ to that line. There are only six other lines, so this is a matter of enumeration.
For question (ii) it suffices to show that for any point, there is a symmetry taking the point $\{1\}$ to that point. There are only six other points, so this is a matter of enumeration.
As for the 24 symmetries fixing a point; again for concreteness sake let's say the point $\{1\}$ is fixed. Then $\{1\}$, $\{2\}$ and $\{4\}$ are not on a line. Let $\{p\}$ and $\{q\}$ be such that $\{1\}$, $\{p\}$ and $\{q\}$ are not on a line. Show that there exists a symmetry fixing $\{1\}$ and mapping $\{2\}$ to $\{p\}$ and $\{4\}$ to $\{q\}$. Show that this determines the symmetry entirely, because symmetries preserve lines.
This is a lot of work and it is very far from insightful. I suggest you follow the more natural approach suggested by Mikhail Katz, which also generalizes much, much better to other finite geometries.
A: To give simple answers
I) you know that there are 168 symmetries in total $168 / 7 =24 $ is the number is symmetries that leave the $1$ on its place (one of these symmetries is the identity) 
Most times there is more than one symmetry that does what you describe 
I haven't counted them but I think that for 2 fixed points there are 4 symmetries that leaves two points fixed but $8$ if you also count the symmetries that exchanges the points.
For symmetries that leave a line fixed a symmetry that swaps  points on the line does count as symmetry that fixes the line 
Hopes this helps
Was puzzling just for a start enumerate (write them all down)  all symmetries where point $1$ stays on the  first spot
Then all where point $3$ stays on the third spot
Looks a lot of tedious work but is quite simple and enlightening 
Then for symmetries that have some qualities you can just check the ones in your list and count them
For example symmetries that fix two lines: 
Suppose line 123 and 356
point 3 is then fixed 
Just count in your list of symmetries where a 3 is on the third spot ,the ones where on the first second fifth and sixth spots are a 1 2 5 and a 6 .
Be aware you will run into some philosophical questions  and there is the learning really happens
Good luck
